1134. Armstrong Number π
Description
Given an integer n
, return true
if and only if it is an Armstrong number.
The k
-digit number n
is an Armstrong number if and only if the kth
power of each digit sums to n
.
Example 1:
Input: n = 153 Output: true Explanation: 153 is a 3-digit number, and 153 = 13 + 53 + 33.
Example 2:
Input: n = 123 Output: false Explanation: 123 is a 3-digit number, and 123 != 13 + 23 + 33 = 36.
Constraints:
1 <= n <= 108
Solutions
Solution 1: Simulation
We can first calculate the number of digits $k$, then calculate the sum $s$ of the $k$th power of each digit, and finally check whether $s$ equals $n$.
The time complexity is $O(\log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the given number.
1 2 3 4 5 6 7 8 |
|
1 2 3 4 5 6 7 8 9 10 |
|
1 2 3 4 5 6 7 8 9 10 11 |
|
1 2 3 4 5 6 7 8 9 10 11 |
|
1 2 3 4 5 6 7 8 |
|
1 2 3 4 5 6 7 8 9 10 11 12 |
|